Re: bug in seq command?

[Andrew D Jewell]

Please clear up my misunderstanding, because it seems to me that the 
fix is indeed trivial.

Note 1) The million below should, of course, be a parameter
Note 2) I am not allowed to submit patches, so the below must be 
taken only as a suggestion.


*** seq.c~      Wed Dec  1 23:45:17 2004
--- seq.c       Sun Jan 23 07:55:07 2005
***************
*** 182,192 ****
  print_numbers (const char *fmt)
  {
    double i;

    for (i = 0; /* empty */; i++)
      {
        double x = first + i * step;
!       if (step < 0 ? x < last : last < x)
        break;
        if (i)
        fputs (separator, stdout);
--- 182,194 ----
  print_numbers (const char *fmt)
  {
    double i;
+   double delta = step/1000000.0;
+   double nlast = last + delta;

    for (i = 0; /* empty */; i++)
      {
        double x = first + i * step;
!       if (step < 0 ? x < nlast : nlast < x)
        break;
        if (i)
        fputs (separator, stdout);



At 10:15 PM -0800 1/21/05, Paul Eggert wrote:
> "Luis A. Florit" <address@hidden> writes:
>
>  > the output of the command:  seq -w 1 0.2 4
> >  is the sequence
> >
> >         1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8
> >
> >  Why there is no 4.0 in the last sequence?
>
> Floating point roundoff error.  It's documented in the manual.
>
> If you can submit a fix, please do!  It would improve seq.
> (Warning: it won't be trivial.)
>
>
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